Describing Motion Around Us · Lesson 8 of 10
Kinematic Equations for Motion in a Straight Line with Constant Acceleration
“Three equations walk into a road problem and only one leaves with the unknown.”
• Relate displacement, time, initial velocity, final velocity and acceleration. • Derive the primary equations from acceleration and graph area. • Use elimination to obtain the equation without time. • Select an equation from known and unknown quantities. • Apply signs consistently in braking problems. • Relate velocity to reaction and braking distances.
A driver sees an obstacle and wants to know whether the vehicle can stop in time. A graph can answer the question, but equations allow the same motion to be predicted directly. For straight-line motion with constant acceleration, five quantities are connected in a compact mathematical system.
| Symbol | Physical quantity |
|---|---|
| s | Displacement |
| t | Time interval |
| u | Initial velocity |
| v | Final velocity |
| a | Constant acceleration |
Start with the definition of acceleration over an interval beginning at zero: (a=(v-u)/t). Multiply by (t) and rearrange. This produces the first equation, which relates velocity and time.
On a velocity-time graph, displacement is the area under the line. When initial velocity is (u), the area can be separated into a rectangle of area (ut) and a triangle of area (½×t×(v-u)). Since (v-u=at), the triangular area becomes (½at^2).
To obtain an equation without time, write (t=(v-u)/a) from the first equation and substitute it into the displacement equation. Simplifying gives a relationship among velocity, acceleration and displacement.
| If this quantity is not needed | Choose |
|---|---|
| Displacement | (v=u+at) |
| Final velocity | (s=ut+½at^2) |
| Time | (v^2=u^2+2as) |
Problem
A car starts from rest and reaches 24 m s⁻¹ in 6 s. Find acceleration and displacement.
- 1.List values: (u=0), (v=24 m s⁻¹), (t=6 s).
- 2.Acceleration: (a=(v-u)/t=(24-0)/6=4 m s⁻²).
- 3.For displacement, use (s=ut+½at^2).
- 4.(s=0×6+½×4×6^2).
- 5.(s=2×36=72 m).
- 6.Check using average velocity for constant acceleration: ((0+24)/2=12 m s⁻¹), and (12×6=72 m).
Problem
A motorbike has initial velocity 28 m s⁻¹ and stops after 98 m. Find acceleration and stopping time.
- 1.Choose forward as positive: (u=28 m s⁻¹), (v=0), (s=98 m).
- 2.Time is not known, so use (v^2=u^2+2as).
- 3.(0=28^2+2a(98)).
- 4.(0=784+196a), so (a=-4 m s⁻²).
- 5.Now use (v=u+at): (0=28-4t).
- 6.Therefore, (t=7 s).
Problem
A car brakes with acceleration (-4 m s⁻²). Compare stopping distances from 54 km h⁻¹ and 108 km h⁻¹.
- 1.Convert velocities: 54 km h⁻¹ = 15 m s⁻¹ and 108 km h⁻¹ = 30 m s⁻¹.
- 2.Use (v^2=u^2+2as) with (v=0).
- 3.For 15 m s⁻¹: (0=225-8s), so (s=28.125 m).
- 4.For 30 m s⁻¹: (0=900-8s), so (s=112.5 m).
- 5.Doubling initial velocity makes the braking distance four times as large under the same acceleration.
Bridging Science and Society
Total stopping distance contains reaction distance and braking distance. During reaction time, the vehicle approximately continues at its initial velocity. Braking distance then depends on velocity, road grip, tyre condition and braking acceleration. Wet surfaces, worn tyres, poor visibility and delayed reaction can all increase the required safe separation.
Problem
A bus travels at 36 km h⁻¹. An obstacle is 30 m ahead. Reaction time is 0.5 s and braking acceleration magnitude is 2.5 m s⁻². Determine whether it stops in time.
- 1.Convert velocity: (36 km h⁻¹=10 m s⁻¹).
- 2.Reaction distance = (ut=10×0.5=5 m).
- 3.After braking begins, use (v^2=u^2+2as) with (v=0) and (a=-2.5 m s⁻²).
- 4.(0=100-5s), so braking distance = 20 m.
- 5.Total stopping distance = (5+20=25 m).
- 6.Because 25 m is less than 30 m, the bus stops 5 m before the obstacle under the stated conditions.
Communication systems that allow nearby vehicles to exchange warnings may reduce reaction delay, but they do not remove the need for suitable speed, safe spacing, maintained tyres and attentive driving.
Note
The equations in this lesson apply only when acceleration is constant. For motion in both directions, the signs of (u), (v), (a) and (s) carry directional meaning. Never replace signed displacement with total distance unless the object moves in one direction throughout.
State the positive direction, convert units, verify constant acceleration and distinguish displacement from distance before substituting values.
Quiz
Which statement is a key takeaway from this lesson?
Which additional statement is also a key takeaway from this lesson?
Which further statement is also a key takeaway from this lesson?
Which another statement is also a key takeaway from this lesson?
Which final statement is also a key takeaway from this lesson?
Practice Problems
- A body starts from rest with acceleration 3 m s⁻² for 5 s. Find final velocity and displacement.
- A car moving at 18 m s⁻¹ stops with acceleration −3 m s⁻². Find stopping time and distance.
- Explain why doubling velocity can quadruple braking distance.
- Find total stopping distance at 12 m s⁻¹ with reaction time 0.6 s and braking acceleration −4 m s⁻².
Key Takeaways
• Three equations connect five motion quantities under constant acceleration. • Equation choice depends on known and unknown quantities. • Signs encode direction. • Braking distance grows with the square of initial velocity for fixed braking acceleration. • Total stopping distance includes reaction and braking distances.